Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{10(3k + 1)}{4k} \times \dfrac{-3}{2(3k + 1)} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ 10(3k + 1) \times -3 } { 4k \times 2(3k + 1) } $ $ z = \dfrac{-30(3k + 1)}{8k(3k + 1)} $ We can cancel the $3k + 1$ so long as $3k + 1 \neq 0$ Therefore $k \neq -\dfrac{1}{3}$ $z = \dfrac{-30 \cancel{(3k + 1})}{8k \cancel{(3k + 1)}} = -\dfrac{30}{8k} = -\dfrac{15}{4k} $